welcome redditors!to snoo-finity ... and beyond!
Welcome to neels11's page.
Contributor score: 1

Comments ...

Subcomments ...

How do you distinguish this from testicular torsion? Is it just because it started in the left flank?

neels11  and there's no mass in the scrotum, whereas testicular torsion will have that "bag of worms" feel (along with a lack of cremaster reflex) testicular torsion usually happens in a younger age group +1  
medpsychosis  @neels11 I would like to clarify a piece of information. I believe you are confusing Varicocele with Testicular Torsion. Varicocele will present with "bag of worms" feeling. While the absence of cremasteric reflex is a sign of testicular torsion. +1  
johnson  This is the classic "loin to groin pain" of nephrolithiasis. +  

submitted by karljeon(30),

I don't know if there is an equation for this, but I basically pumped out every division across the table to get ~5% on average.

Here they are: 400 / 6,000 = 0.067 250 / 5,600 = 0.045 300 / 5,350 = 0.056 300 / 5,050 = 0.059 250 / 4,800 = 0.052

The average of these %s for all the years = 5.58%. So that's close enough to 5%.

seagull  good work. I found this question annoying and gave up doing those considering the amount of time we are given. +3  
vshummy  Well just don’t include the intake year... because that messed me up.. +4  
_yeetmasterflex  How would we have known not to include the intake year? From average **annual** incidence? +  
lamhtu  Do not include intake year because the question stem is asking average annual incidence. The 4000 positives at intake could have acquired HIV whenever, not just in the last year. +1  
neels11  literally didn't think there was an actual way to figure this out. but my thought process was: okay incidence means NEW cases. so the annual average at the end of 5 years would be: (# of NEW people that tested positive at the end of year 5) / (# of people at that were at risk at the beginning of year 5) <--- aka at the end of year 4 250/5050 = 4.95% also if you look at year 5: you'll see that the at risk population is 4800 when 300 new cases were found the year before. 5050 at the end of year 4 MINUS the 300 new cases at the end of year 4 should give you 4750 as the new population at risk. but notice that end of year 5 we have 4800. idk if that means 50 people were false positives before or 50 people were added but in incidence births/death/etc don't matter it's kind of like UWORLD ID 1270. assuming average annual incidence is the same as cumulative incidence this was just a bunch of word vomit. sorry if it was unbearable to follow +